tom.stoer said:
No, you will never learn
why xyz is true ...
The Pauli principle "no two fermions can occupy identical quantum states" follows from the antisymmetry of fermions wave functions, which follows from the spin-statistics theorem, which follows from Lorentz invariance and some other assumptions (there are some exotic counteraxemples which become important when some assumptions do no longer apply!), which follow from ...
The are no ultimate justifications in physics.
Anyway - it makes sense to start with
http://en.wikipedia.org/wiki/Spin-statistics_theorem
Well, less complicated than the spin-statistics theorem is to understand the fact that there should be symmetry under an exchange of particles. I don't remember the argument in full, but it went something like this...
If \Psi(x_1, x_2) is the probability amplitude for particle 1 to be at position x_1 and particle 2 to be at position x_2, then, since particles of the same type are indistinguishable, we know that
\Psi(x_1, x_2) and \Psi(x_2, x_1)
have the same physical content. That doesn't mean that they are equal, because wave functions have an arbitrary phase associated that doesn't make any observable difference. So switching two identical particles could possibly introduce a phase change. So we write:
\Psi(x_1, x_2) = e^{i \alpha} \Psi(x_2, x_1)
where \alpha is a phase difference. Now, we can reason that if we switched the particles
again, we have to get exactly back to where we started. So we reason that:
\Psi(x_1, x_2) = e^{2 i \alpha} \Psi(x_1, x_2)
which implies that
e^{2 i \alpha} = 1
So \alpha = 0 or \pi.
This is from memory, so there might be some subtle points that I've forgotten, but something along this lines is supposed to show that particles must either be Bosons (exchanging two identical ones makes no difference) or Fermions (exchanging two identical ones changes the sign of the wave function).
